I have a query where I am trying to obtain the id and name for a board item based on a column value. I am trying to pass the variable $name to the query like this:
$name = “Test Practice Full Service”;
$query = 'query ($name: String!){
items_by_column_values (board_id: 3873753003, column_id: "name", column_value: $name) {
id
name
}
}';
However, I get an error message:
Variable $name of type String! was provided invalid value
what am I doing wrong here, and how should I structure this query so that I can pass the $name variable into the query?